 is going to be a little with Paley family, OK? And we're going to further suppose that there exists constants. Let's see, what am I calling them, c0 and delta. And for each dietic cube, q, a function bq such that the following is true. We have the following three conditions, OK? First condition is that bq is in L2. And the L2, while the square of the L2 norm, is controlled by the measure of q, OK? And you might think, is that a little unnatural? Maybe if we're talking about local things, maybe the bq should be supported near q. In fact, you can actually think of it that way. It's actually equivalent to formulate the theorem that way, right? So you could just as well think of bq as being supported in, let's say, 5 times q. OK, but it's kind of convenient to do it this way. It simplifies the proof slightly. There's just one last step we have to do, OK? Second thing is this sort of weak equitivity condition, which is that the average of q, the average of bq, on its own cube q with absolute values on the outside or taking the modulus, because this could be complex value, is bounded below by delta, OK? All right, and the third thing is that, well, it's sort of like a Carlson condition, except not really because the measure itself is depending on q. What we're going to assume is that if we integrate over the Carlson region associated to q, theta t applied to the testing function for that cube q, that's also going to be less than or equal to c0 times the measure of q, OK? So before we prove it, maybe a couple of remarks are in order, OK? Notice that if bq is the same, it's just one function b, one accredited function b, or it's even pseudo-creditive, for all q, then this is just some theorem. Then this is theorem 4.4, because if this measure here is not depending on q, then this is a Carlson measure estimate, OK? Yes? Oh, thank you. Yeah, I guess I should. OK, we'll come back to the remarks in a second. Thank you. Thank you very much. It's always important to give the conclusion of your theorem. You probably already guessed what the conclusion is. The conclusion is that the theta t gives rise to a bound of square function, OK? So the square of the L2 norm of the square function associated to theta t is bounded by the square of the L2 norm of f, where, of course, c is going to depend on all the usual stuff. C is going to depend on dimension, a little with Paley estimates, the constant c0, the constant delta. I guess that's all the constants that I have, OK? All right, and then one more remark, which is that condition 2 is a sort of weak pseudo-accurativity condition, right? Because if the average of this function over every cube, cube, every dyadic cube, cube had this property, that would be dyadic pseudo-accurativity. But the point is that we're only assuming the securitivity condition for Bq integrated on its own cube, q, not on other cubes, OK? Yes, if Bq is equal to some constant function, not constant function, but to some globally defined function, B for all q, right? That's when you have to do the localization thing. I glossed over that. Yeah, remember I was saying that you could equivalently formulate this with a localization here. Yeah, thank you, you're right. That should be clarified, OK? Modulo localization, you're exactly right, OK? And in fact, that also brings up another point, was that notice that, I mean, because you can't, I mean, I haven't explained why, but you can localize this. And since you can localize it, somehow this is weaker than saying that the B is in L infinity, right? You're only insisting on L2, not L infinity, OK? All right, and the last thing is that the last remark is that this is sort of a weak dyadic pseudo-accretivity. We assume non-degeneracy of Bq averaged only over its own cube q and not on other cubes, OK? All right, nonetheless, we're going to find out that there are still many, many cubes on which it is accretive, OK? All right, so how do we prove this? All right, so what we want to do is to use, again, we're going to reduce to the T of 1 theorem for square functions. That was theorem, was that theorem 3.1? Yes, theorem 3.1, OK? In other words, we need to show that the integral 0 to the length of q, the soup on q, I'm being too small here, the soup on q of 1 over the measure of q, integral 0 to the length of q, integral on q, theta T of 1, squared dx dt over T, is less than some uniform constant, all right? We boil us down to that because we have a little wood-Paley kernel for our theta T's, all right? But now, here's where we're going to use that John Nirenberg theorem, or John Nirenberg lemma, OK? So by the Jn lemma that we just proved a moment ago, it's actually enough to show that there exist constants c1 and a double positive. And for each q, dyadic cube, a family consisting of non-overlapping dyadic subcubes of q, such that the complement of the qj's, which is this set that we call the eq, is at least an eta portion of the measure of q. And such that this full Carlson measure estimate that I wrote down is replaced, where instead of integrating over all of the Carlson box rq, we only integrate over eq star, OK? So that's what we need to show, right? And then if we can do that, then we invoke the John Nirenberg. And this turns into that, and we're done, OK? All right, so what are we going to do? OK, so we're going to do this by a stopping time argument. OK, so the first thing we're going to do just for convenience, ah, I've erased things now. But notice that by rescaling or by normalizing bq, we can assume, in fact, that the average on q of bq is precisely 1, right? You just divide bq by its average on q, OK? And the price you pay is that then the constant c0 becomes c0 over delta squared, if you think how the normalization would go then. Then the constant c0 goes to c0 over delta squared. But for simplicity of notation, we'll just again call this, call it c0, OK? But remember, for now on, when you see a c0, it's really c0 over delta squared, OK? All right, so we're starting off with this. And we're going to subdivide q, diatically, and we stop the first time we get a qq. We get average on q prime or q prime sub-cube of q, where the real part of the average of bq is less than or equal to 1 half. That's when we stop, OK? All right, and we let the family f then simply be the maximal cubes that we get with respect to this property, right? The cubes that are maximal with respect to that property, the cubes that were we stopped the first time, OK? Our cubes that are maximal are the dietic sub-cubes of q that are maximal with respect to the property that, let's call it dagger, that the real part of the average on qj of bq is less than or equal to 1 half, OK? All right? And now we just need to show that these guys are the qj's which let us do these things, OK? So first of all, well, maybe let me see it this way. So the claim then is that for this family f, we have the following two things. That a, eq, which of course, again, is the complement of the qj's, is at least such that eq is bigger than or equal to a that times the measure of q, OK? OK, and b, such that for x and t in the associated eq star, remember which is rq minus the union of the rqj's, we have that the real part of the dyadic averaging operator of bq at x for that x and t is going to be bigger than or equal to 1 half, all right? OK, so let's see what that is, OK? Of course, a is just a restatement of this, all right? And b is what's going to allow us to prove that, OK? So remember we've normalized so that the average of bq and q is exactly 1. So we have measure of q equals integral on q of bq, which is going to be then we're splitting that into the integral on eq plus the sum of the integrals on the stopping time cubes, qj, all right? And notice, OK, since this average was equal to 1, which is a real number, we can take real parts and that doesn't change anything. So we can actually take the real part here and the real part here, OK? All right, and then here we use Cauchy-Schwarz. This is going to be less than or equal to square to the measure of bq times the L2 norm of bq. And this is going to be what? Well, remember what was the stopping time criterion? The real part of the integral on qj is bounded by 1 half times the measure of qj. So this is less than or equal to 1 half times the sum of the measures that qj is. But these are non-overlapping dyadic subcubes of q. You sum them up. At worst, it's q. It's actually less than q. But at worst, this gives you the measure of q with a half, OK? And this, if you remember, we had an estimate for the square of the L2 norm of bq. So when you take the L2 norm, you get that. And this, of course, is trivially less than or equal to 1 half times the measure of q. And of course, you hide that bit, OK? So you hide that bit. And we have now 1 half measure of q on the left-hand side. And then you divide through by this. And you get 1 half, measure of q to the 1 half is less than or equal to square root of c0, measure of vq to the 1 half, square both sides. And you get we obtain a with a to being what? 1 over 4 c0, right? OK, so what about b? Well, b is essentially automatic just by definition, OK? So remember what is the, recall what is the definition of the dyadic averaging operator. This is the average over the unique half open cube q of xt, where q of xt is containing x and has side length at least t. So that it's a minimal one with side length at least t. So that length of q of x over t over 2 is less than t is less than or equal to the length of q of xt. Sorry, that got kind of cramped, but all right? And if you think about that, what that means is that consequently, then by construction, if xt is in eq star, then the q of xt that we get is not contained in any qj. You can see that by just a little momentary reflection on the geometry of the situation. And what does that do for us? Well, remember, these guys were the maximal cubes with respect to this property. So if you've got a cube q of xt that's not contained in any of these guys, then that's a cube where in the stomping time construction you have not stopped yet. And so the opposite inequality to this holds, OK? So hence, have the opposite inequality to this thing we call dagger. And therefore, what do we have? We have that delta, not delta, a half. A half is going to be less than or equal to the real part of at bq of x for all xt in eq star, OK? All right, and that proves b, OK? All right, so the last thing we need to do, remember a, which we've already proved, gives the ampleness of the complement of qj's. And the last thing we need to do is somehow use this to get the control of theta t of 1 in the sawtooth, OK? So consequently, for x and t in eq star, we have that theta t of 1 of x will be less than or equal to twice theta t of 1 of x times at bq of x, OK? All right, which means that therefore, when we integrate on eq star theta t of 1 squared dx dt over t, this will be less than or equal to 4 times the same integral. And at that point, we're going to go ahead and integrate over all of our q, because this guy's work is done once we've got this inequality. And we have theta t of 1 times at bq squared dx dt over t, OK? And now we use a trick that we've used a number of times at this point, this little trick of quite from my error, OK? What we're going to do is write theta t of 1 times at of b. We're going to write it as itself minus theta t plus theta t, OK? It's the familiar thing that we've been doing. So this will be theta t of 1 of x times at minus theta t applied to bq. Oh, I guess I need more parentheses. And then applied to n, and then evaluated at x. And then we have plus theta t bq of x. And then what happens? Well, then we have to look at the contribution of each of these guys in here, all right? And the contribution of this guy is OK by hypothesis. That was hypothesis 3. It said precisely that theta t of bq integrated over rq with respect to this measure gives you the bound that you want, OK? And the other part, we're going to call this rt of bq. Of course, there's the usual thing that by construction, since at are averaging operators, this at of 1 is 1, we therefore have rt of 1 is 0, OK? And then that last guy we handle in the usual way. We handle rt in the usual way. So therefore, it's enough to show that we have quasi-arthogonality up to some constant, that there's some positive alpha. Oh, maybe I shouldn't use alpha, because alpha's the little bit exponent for some beta positive. Right, because once we have this, if that tells you that gives you bounded square functions, you pick up the l2 norm of bq, but then you use, you get a bound in terms of, so you get, and then in that case, you get a bound on the order of the square of the l2 norm of bq. But remember, by hypothesis 1, that's controlled by the thing that you want, and then you're done. All right, so this part, I'll leave as a bit of an exercise. It's, you don't quite have, because of this at, it doesn't quite have the smoothness, the standard little wood-paley smoothness, but it does satisfy this weak sort of l1 condition that you work with in exercise 2. And then you get this, and then you're done, OK? So that's it. Thank you for your attention. Any questions? Any questions? All right, well, thanks for all your attention. Have a good rest of your time here.